Answer
(a) $499$ terms are needed.
(b) ${K_{499}} \approx 0.915965$ approximates the Catalan's constant $K = \mathop \sum \limits_{k = 0}^\infty \dfrac{{{{\left( { - 1} \right)}^k}}}{{{{\left( {2k + 1} \right)}^2}}} \approx 0.915966$ to within an error of less than ${10^{ - 6}}$.
Work Step by Step
(a) We have $K = \mathop \sum \limits_{k = 0}^\infty \dfrac{{{{\left( { - 1} \right)}^k}}}{{{{\left( {2k + 1} \right)}^2}}}$, where the positive terms are ${b_k} = \dfrac{1}{{{{\left( {2k + 1} \right)}^2}}}$.
By Eq. (2) in Theorem 3 of Section 11.4:
$\left| {K - {K_n}} \right| \lt {b_{n + 1}}$
We choose $n$ so that
$\left| {K - {K_n}} \right| \lt {b_{n + 1}} \lt {10^{ - 6}}$
Thus,
$\dfrac{1}{{{{\left( {2n + 3} \right)}^2}}} \lt {10^{ - 6}}$
${\left( {2n + 3} \right)^2} \gt {10^6}$
We solve this inequality and obtain $n \gt 498.5$.
So, we choose $N=499$.
Thus, we need $499$ terms of the series to calculate $K$ with an error of less than ${10^{ - 6}}$.
(b) Using a computer algebra system, we compute:
${K_{499}} = \mathop \sum \limits_{k = 0}^{499} \dfrac{{{{\left( { - 1} \right)}^k}}}{{{{\left( {2k + 1} \right)}^2}}} \approx 0.915965$, ${\ \ \ \ \ }$ $K = \mathop \sum \limits_{k = 0}^\infty \dfrac{{{{\left( { - 1} \right)}^k}}}{{{{\left( {2k + 1} \right)}^2}}} \approx 0.915966$
$\left| {K - {K_{499}}} \right| \approx 4.99999 \times {10^{ - 7}} \lt {10^{ - 6}}$
Thus, ${K_{499}} \approx 0.915965$ approximates the Catalan's constant $K = \mathop \sum \limits_{k = 0}^\infty \dfrac{{{{\left( { - 1} \right)}^k}}}{{{{\left( {2k + 1} \right)}^2}}} \approx 0.915966$ to within an error of less than ${10^{ - 6}}$.
